The charge density of a spherical conductor of radius 50 cm is 8-85 - 10 - 6 C - m 2- The electric field intensity in V-m at a point close to the surface will be -

Given, #10; #10;Radius of sphere, r=0.5 m #10; #10;Charge density, qA=8.85× #10;10^ 6 Cm^ 2 #10; #10;From gauss law, #10; #10;El ...

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Standard XII

Physics

Displacement Current

The charge de...

Question

The charge density of a spherical conductor of radius $50 c m$ is $8.85 \times 10^{- 6} C / m^{2}$ . The electric field intensity in $V / m$ at a point close to the surface will be :-

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Solution

Given,

Radius of sphere, $r = 0.5 m$

Charge density, $\frac{q}{A} = 8.85 \times 10^{- 6} C m^{- 2}$

From gauss law,

Electric field intensity, $E = \frac{q}{A ε_{o}} = \frac{8 \times 10^{- 6}}{8 \times 10^{- 12}} = 10^{6} V / m$

Hence electric field intensity is $10^{6} V / m$

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The charge density of a spherical conductor of radius 50cm is 8.85×10−6C/m2. The electric field intensity in V/m at a point close to the surface will be :-

Given, Radius of sphere,r=0.5m Charge density, qA=8.85×10−6Cm−2 From gauss law, Electric field intensity, E=qAεo=8×10−68×10−12=106V/m Hence electric field intensity is 106V/m

The charge density of a spherical conductor of radius $50 c m$ is $8.85 \times 10^{- 6} C / m^{2}$ . The electric field intensity in $V / m$ at a point close to the surface will be :-

Given,

Radius of sphere, $r = 0.5 m$

Charge density, $\frac{q}{A} = 8.85 \times 10^{- 6} C m^{- 2}$

From gauss law,

Electric field intensity, $E = \frac{q}{A ε_{o}} = \frac{8 \times 10^{- 6}}{8 \times 10^{- 12}} = 10^{6} V / m$

Hence electric field intensity is $10^{6} V / m$